# Group Homomorphism Preserves Identity

## Theorem

Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.

Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group homomorphism.

Let:

$e_G$ be the identity of $G$
$e_H$ be the identity of $H$.

Then:

$\map \phi {e_G} = e_H$

## Proof 1

 $\ds \map \phi {e_G}$ $=$ $\ds \map \phi {e_G \circ e_G}$ Definition of Identity Element $\ds$ $=$ $\ds \map \phi {e_G} * \map \phi {e_G}$ Definition of Morphism Property

That is:

 $\ds \map \phi {e_G} * e_H$ $=$ $\ds \map \phi {e_G} * \map \phi {e_G}$ Definition of Identity Element $\ds \leadsto \ \$ $\ds e_H$ $=$ $\ds \map \phi {e_G}$ Cancellation Laws

$\blacksquare$

## Proof 2

A direct application of Homomorphism to Group Preserves Identity.

$\blacksquare$

## Proof 3

From Group Homomorphism of Product with Inverse, we have:

$\forall x, y \in G: \map \phi {x \circ y^{-1} } = \map \phi x * \paren {\map \phi y}^{-1}$

Putting $x = y$ we have:

 $\ds \map \phi {e_G}$ $=$ $\ds \map \phi {x \circ x^{-1} }$ $\ds$ $=$ $\ds \map \phi x * \paren {\map \phi x}^{-1}$ $\ds$ $=$ $\ds e_H$

$\blacksquare$